PHYS 424 Notes

Let us suppose we have a vector space of functions with a basis |n> or, equivalently, y_n(x)

The solutions of a Hamiltonian H will do for the basis, provided that the solutions are normalizable and complete. The inner product will be < n | m> in abstract notation or

ò y_n(x)^* y_m(x) dx

in more explicit notation. We have seen that we sometimes need expressions like

x y_m(x) º x | m > . We can expand x y_m in terms of the basis to get

x |m > = S_n | n > a_nm.                   [1]

Taking the inner product with y_q (x) on the left-hand side of [1] gives

ò y_q(x)^* x y_m(x) dx = <q |[ x | m>] = <q | x | m>

where the last form is just a simpler way of writing the second, not a different expression. On the right-hand side of [1] we get

< q | S_n | n > a_nm = S_n< q | n > a_nm

The orthonormality of the basis gives

< q | n > = d_qn

so that, putting the two sides of [1] together

< q | x | m > = a_qm

We can, therefore, determine the coefficients of the expansion of x | m > in terms of integrations involving the basis functions. The quantity < q | x | m >, since it is a set of numbers with two subscripts labelling the members of the set, is generally referred to as the "pm matrix element" of x.

The matrix elements of the Hamiltonian are special, since the Hamiltonian determines the usual basis for doing quantum mechanics. The Schroedinger equation gives

H | n> = E_n | n > so that

<m | H | n > = E_n d _mn

which means that the matrix version of the Hamiltonian is diagonal in the normal basis . We could in fact seek the y_n(x) by looking for the basis in which the Hamiltonian is diagonal. This is generally called "diagonalizing the Hamiltonian." In this context, the y_n are called eigenvectors and the E_n are called eigenvalues.

For an operator whose matrix is a particular type called "Hermitian," the eigenvalues are known to be real and hence suitable numbers for physical quantities. For finite Hermitian matrices, the eigenvectors are orthonormal and complete, and for many infinite matrices the same is true. Hence we need to define Hermitian operators (matrices) and show that useful physical quantities are represented by Hermitian operators.

For this purpose we need, briefly, a more explicit notation for the scalar product. The expression is more awkward than < m | n>, but useful:

<m|n> = ( |m> , |n> )= (|n> , |m> )^*.

In this notation, and using T for an arbitrary operator T_mn = (|m>, T |n>) . We define (T^hc)_mn = T_nm^* = ( |n> , T |m> )^* = ( T |m> , |n>). Collecting the outside pieces, the Hermitian conjugate of an operator is given by

In the case of x_op, the condition is

ò dx y_m^* x y_n = ò dx (x y_m)^* y_n

which is obviously true. For momentum, the condition is

ò dx y_m^* [-i hbar (d/dx)] y_n = ò dx [-i hbar (d/dx) y_m]^* y_n

which is true after an integration by parts provided that no surface terms show up (and provided the operator does not carry us outside the working vector space, a caveat we should also apply to x_op). Thus ordinarily both x and p are Hermitian operators. The importance of this property is our next task.

The time-independent Schroedinger equation is

Hy_n = Ey_n

We can generalize this equation to an arbitrary operator T so that

Ty_a = l y_a   or   T_op | a> = l | a>

In terms of a basis in which |a> = S_m a_m|e_m>,

we have

S_ma_m T_op | m > = l S_m a_m | m >

S_ma_m < n | T_op | m > = l S_m a_m <n | m >

S_mT_nm a_m = S_m l d_nm a_m

T a = l a       or (T - l1) a = 0 ,

where 0 is a column matrix whose entries are all zeroes. This set of equations has a nonzero solution if and only if

det (T - l1) = 0,

which for an n x n matrix yields an nth order equation for l. There are n complex solutions for l, not all necessarily distinct. There is a column matrix a corresponding to each l. If, as will normally be true in physics, T is Hermitian [since it is obtained from combinations of the Hermitian x and p], l is real, the eigenvectors a can be chosen orthogonal, the eigenvectors corresponding to different l's are automatically orthogonal, and the eigenvectors span the vector space unless possibly if the space is infinite.

I will work out an example of the determination of the eigenvalues and eigenvectors of a Hermitian matrix. Note that the example in Griffiths is a non-Hermitian matrix, and non-Hermitian matrices are more difficult and very unusual. Choose

       (0  -i   0)                  | -l  -i   0 |
 T =   (i   0  -i)     |T - l I | = |  i  -l  -i |
       (0   i   0)                  |  0   i  -l |

which yields a characteristic equation of

-l³ + 2 l = 0 => l = 0, ± Ö2.

There is a completely mechanical way to determine the corresponding eigenvectors. The cofactors of the first row of T - l I are

(+)(l²-1),     (-)(-il),     (+)(-1)

or, substituting the values of l,

(1, iÖ2, -1) for l₁ = +Ö2,
(-1, 0, -1) for l₂=0
(1, -iÖ2, -1) for l₃ = -Ö2.

The corresponding normalized eigenvectors are

a⁽¹⁾ = (1/2) (1, iÖ2, -1)^tr

a⁽²⁾ = (1/Ö2) (1, 0, 1)^tr

a⁽³⁾ = (1/2) (1, -iÖ2, -1)^tr

If, for any value of l, all the components of the eigenvector vanish, you must use the cofactors of a different row to calculate the eigenvector [for that value of l only].

For any Hermitian matrix H

• The eigenvectors corresponding to different eigenvalues are orthogonal, and the eigenvectors corresponding to equal eigenvalues may be chosen orthogonal.

• For a finite matrix, the eigenvectors are complete. the same is true for most or all infinite matrices of physical interest.

Function Spaces -- Miscellany

• We will do Legendre polynomials later.

• Hilbert space - name for an infinite-dimensional complete vector space.

• Eigenfunctions g_x0(x) of x_op and f_k(x) of p_op:
  x_opd(x-x₀) = x₀d(x-x₀)
  (-i hbar d/dx)[(2p) ^-1/2 e ^{i k x}] = hbar k [(2p) ^-1/2 e ^{i k x}]

  which satisfy

  <x'₀ | x₀ > = d(x'₀ - x₀)

  <k' | k > = d(k' - k)

  These functions are not normalizable and therefore are not part of the Hilbert space. "Normalizing" to a Dirac delta is the best we can do. The eigenvectors are useful even though they are not members of the Hilbert space.